How much must you raise the temperature of 1.00m^3 of water so that its volume expands by 0.0100

For an ideal gas, product of pressure and volume equals n times R times T.

Mathematically,        PV = nRT

where        P = pressure

                V = volume

                n = number of moles

                R = gas constant

                T = temperature

Also, it is known that in 1 kpa there are 0.0098 atm. So, 100.0 kpa equals 0.98 atm. And 275 ml equals 0.275 l. Therefore, calculate temperature as follows.

                       PV = nRT

                     T =  

Thus, we can conclude that temperature for the given problem is 335k .

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